1. Rotational Motion and Angular Momentum
Chapter Opener. Caption: You too can experience rapid rotation—if your stomach can take the high angular velocity and centripetal acceleration of some of the faster
amusement park rides. If not, try the slower merry-go-round or Ferris wheel. Rotating carnival rides have rotational kinetic energy as well as angular momentum. Angular acceleration is produced by a net torque, and rotating objects have rotational kinetic energy.

2. Angular Quantities
Vector Nature of Angular Quantities
Constant Angular Acceleration
Torque
Vector Cross Product
Rotational Dynamics; Torque and Rotational Inertia
Solving Problems in Rotational Dynamics

3. Determining Moments of Inertia
Rotational Kinetic Energy
Rotational Plus Translational Motion; Rolling
Why Does a Rolling Sphere Slow Down?

4. Angular Momentum—Objects Rotating About a Fixed Axis
Angular Momentum of a Particle
Angular Momentum and Torque for a System of Particles; General Motion
Angular Momentum and Torque for a Rigid Object

5. Conservation of Angular Momentum
The Spinning Top and Gyroscope
Rotating Frames of Reference; Inertial Forces
The Coriolis Effect

6. Angular Quantities
In purely rotational motion, all points on the object move in circles around the axis of rotation (“O”). The radius of the circle is R. All points on a straight line drawn through the axis move through the same angle in the same time. The angle θ in radians is defined:
where l is the arc length.
Figure Caption: Looking at a wheel that is rotating counterclockwise about an axis through the wheel’s center at O (axis perpendicular to the page). Each point, such as point P, moves in a circular path; l is the distance P travels as the wheel rotates through the angle θ.

7. Angular Quantities Birds of prey—in radians.
A particular bird’s eye can just distinguish objects that subtend an angle no smaller than about 3 x 10-4 rad. (a) How many degrees is this? (b) How small an object can the bird just distinguish when flying at a height of 100 m?
Figure Caption: (a) Example 10–1. (b) For small angles, arc length and the chord length (straight line) are nearly equal. For an angle as large as 15°, the error in making this estimate is only 1%. For larger angles the error increases rapidly.
Solution: a °
b. 3 cm (assuming the arc length and the chord length are the same)

8. Solution:

9. Angular Quantities Angular displacement:
The average angular velocity is defined as the total angular displacement divided by time:
The instantaneous angular velocity:
Figure Caption: A wheel rotates from (a) initial position θ1 to (b) final position θ2. The angular displacement is Δθ = θ2 – θ1.

10. Angular Quantities
The angular acceleration is the rate at which the angular velocity changes with time:
The instantaneous acceleration:

11. Angular Quantities
Every point on a rotating body has an angular velocity ω and a linear velocity v.
They are related:
Figure Caption: A point P on a rotating wheel has a linear velocity v at any moment.

12. Angular Quantities Is the lion faster than the horse?
On a rotating carousel or merry-go-round, one child sits on a horse near the outer edge and another child sits on a lion halfway out from the center. (a) Which child has the greater linear velocity? (b) Which child has the greater angular velocity?
Answer: The horse has a greater linear velocity; the angular velocities are the same.

13. Angular Quantities
Objects farther from the axis of rotation will move faster.
Figure Caption: A wheel rotating uniformly counterclockwise. Two points on the wheel, at distances RA and RB from the center, have the same angular velocity ω because they travel through the same angle θ in the same time interval. But the two points have different linear velocities because they travel different distances in the same time interval. Since RB > RA, then vB > vA (because v = Rω).

14. Angular Quantities
If the angular velocity of a rotating object changes, it has a tangential acceleration:
Even if the angular velocity is constant, each point on the object has a centripetal acceleration:
Figure Caption: On a rotating wheel whose angular speed is increasing, a point P has both tangential and radial (centripetal) components of linear acceleration. (See also Chapter 5.)

15. Angular Quantities
Here is the correspondence between linear and rotational quantities:

16. Angular Quantities Angular and linear velocities and accelerations.
A carousel is initially at rest. At t = 0 it is given a constant angular acceleration α = rad/s2, which increases its angular velocity for 8.0 s. At t = 8.0 s, determine the magnitude of the following quantities: (a) the angular velocity of the carousel; (b) the linear velocity of a child located 2.5 m from the center; (c) the tangential (linear) acceleration of that child; (d) the centripetal acceleration of the child; and (e) the total linear acceleration of the child.
Figure Caption: Example 10–3. The total acceleration vector a = atan + aR, at t = 8.0 s.
Solution: a. The angular velocity increases linearly; at 8.0 s it is 0.48 rad/s.
b. The linear velocity is 1.2 m/s.
c. The tangential acceleration is 0.15 m/s2.
d. The centripetal acceleration at 8.0 s is 0.58 m/s2.
e. The total acceleration is 0.60 m/s2, at an angle of 15° to the radius.

17. Solution:

18. Angular Quantities
The frequency is the number of complete revolutions per second:
Frequencies are measured in hertz:
The period is the time one revolution takes:

19. Angular Quantities Hard drive.
The platter of the hard drive of a computer rotates at 7200 rpm (rpm = revolutions per minute = rev/min). (a) What is the angular velocity (rad/s) of the platter? (b) If the reading head of the drive is located 3.00 cm from the rotation axis, what is the linear speed of the point on the platter just below it? (c) If a single bit requires 0.50 μm of length along the direction of motion, how many bits per second can the writing head write when it is 3.00 cm from the axis?
Solution: a. F = 120 Hz, so the angular velocity is 754 rad/s.
b. V = 22.6 m/s.
c. 45 megabits/s

20. Angular Quantities Given ω as function of time.
A disk of radius R = 3.0 m rotates at an angular velocity ω = ( t) rad/s, where t is in seconds. At the instant t = 2.0 s, determine (a) the angular acceleration, and (b) the speed v and the components of the acceleration a of a point on the edge of the disk.
Solution: a. The angular acceleration is the derivative of the angular velocity: 1.2 rad/s2.
b. V = 12.0 m/s; atan = 3.6 m/s2; aR = 48 m/s2.

21. Vector Nature of Angular Quantities
The angular velocity vector points along the axis of rotation, with the direction given by the right-hand rule. If the direction of the rotation axis does not change, the angular acceleration vector points along it as well.
Figure Caption: (a) Rotating wheel. (b) Right-hand rule for obtaining direction of ω.

22. Constant Angular Acceleration
The equations of motion for constant angular acceleration are the same as those for linear motion, with the substitution of the angular quantities for the linear ones.

23. Constant Angular Acceleration
Centrifuge acceleration.
A centrifuge rotor is accelerated from rest to 20,000 rpm in 30 s. (a) What is its average angular acceleration? (b) Through how many revolutions has the centrifuge rotor turned during its acceleration period, assuming constant angular acceleration?
Solution: a. The final angular velocity is 2100 rad/s, so the acceleration is 70 rad/s2.
b. The total angle is 3.15 x 104 rad, which is 5000 rev.

24. Solution:

25. Torque
To make an object start rotating, a force is needed; the position and direction of the force matter as well.
The perpendicular distance from the axis of rotation to the line along which the force acts is called the lever arm.
Figure Caption: Top view of a door. Applying the same force with different lever arms, RA and RB. If RA = 3RB, then to create the same effect (angular acceleration), FB needs to be three times FA, or FA = 1/3 FB.

26. Torque A longer lever arm is very helpful in rotating objects.
Figure Caption: (a) A tire iron too can have a long lever arm. (b) A plumber can exert greater torque using a wrench with a long lever arm.

27. Torque
Here, the lever arm for FA is the distance from the knob to the hinge; the lever arm for FD is zero; and the lever arm for FC is as shown.
Figure Caption: (a) Forces acting at different angles at the doorknob. (b) The lever arm is defined as the perpendicular distance from the axis of rotation (the hinge) to the line of action of the force.

28. Torque The torque is defined as:
Figure Caption: Torque = R┴ F = RF┴.

29. Torque Torque on a compound wheel.
Two thin disk-shaped wheels, of radii RA = 30 cm and RB = 50 cm, are attached to each other on an axle that passes through the center of each, as shown. Calculate the net torque on this compound wheel due to the two forces shown, each of magnitude 50 N.
Figure Caption: Example 10–7.The torque due to FA tends to accelerate the wheel counterclockwise, whereas the torque due to FB tends to accelerate the wheel clockwise.
Solution: The net torque is RAFA – RBFB sin 60° = -6.7 m·N.

30. Solution:
The torque due to FA tends to accelerate the wheel counterclockwise, whereas the torque due to FB tends to accelerate the wheel clockwise.

31. Vector Cross Product The vector cross product is defined as:
The direction of the cross product is defined by a right-hand rule:
Figure The vector C = A x B is perpendicular to the plane containing A and B; its direction is given by the right-hand rule.

32. Vector Cross Product
The cross product can also be written in determinant form:

33. Vector Cross Product Some properties of the cross product:
Figure The vector B x A equals –A x B.

34. Torque as a Vector
Torque can be defined as the vector product of the force and the vector from the point of action of the force to the axis of rotation:
Figure The torque due to the force F (in the plane of the wheel) starts the wheel rotating counterclockwise so ω and α point out of the page.

35. Torque as a Vector
For a particle, the torque can be defined around a point O:
Here, is the position vector from the particle relative to O.
Figure τ = r x F, where r is the position vector.

36. Torque as a Vector Torque vector.
Suppose the vector is in the xz plane, and is given by = (1.2 m) m) Calculate the torque vector if = (150 N) .
Solution: This can be done by the determinant method; the answer is τ = (180 m·N)j – it is in the y-direction, as expected.

37. Torque and Rotational Inertia
Knowing that , we see that
This is for a single point mass; what about an extended object?
As the angular acceleration is the same for the whole object, we can write: R
Figure Caption: A mass m rotating in a circle of radius R about a fixed point.

38. Torque and Rotational Inertia
The quantity is called the rotational inertia of an object.
The distribution of mass matters here—these two objects have the same mass, but the one on the left has a greater rotational inertia, as so much of its mass is far from the axis of rotation.
Figure Caption: A large-diameter cylinder has greater rotational inertia than one of equal mass but smaller diameter.

39. Torque and Rotational Inertia
The rotational inertia of an object depends not only on its mass distribution but also the location of the axis of rotation—compare (f) and (g), for example.

40. Solving Problems in Rotational Dynamics
Draw a diagram.
Decide what the system comprises.
Draw a free-body diagram for each object under consideration, including all the forces acting on it and where they act.
Find the axis of rotation; calculate the torques around it.

41. Solving Problems in Rotational Dynamics
5. Apply Newton’s second law for rotation. If the rotational inertia is not provided, you need to find it before proceeding with this step.
6. Apply Newton’s second law for translation and other laws and principles as needed.
7. Solve.
8. Check your answer for units and correct order of magnitude.

42. Determining Moments of Inertia
If a physical object is available, the moment of inertia can be measured experimentally.
Otherwise, if the object can be considered to be a continuous distribution of mass, the moment of inertia may be calculated:

43. Determining Moments of Inertia
Cylinder, solid or hollow.
(a) Show that the moment of inertia of a uniform hollow cylinder of inner radius R1, outer radius R2, and mass M, is I = ½ M(R12 + R22), if the rotation axis is through the center along the axis of symmetry. (b) Obtain the moment of inertia for a solid cylinder.
Figure Caption: Determining the moment of inertia of a hollow cylinder.
Solution: a. The cylinder can be thought of as consisting of a series of thin rings of thickness dR located a distance R from the center. The mass of each ring is dm = ρdV = 2πρhR dR. The moment of inertia I = ∫R2 dm = πρh/2 (R24 - R14). Factoring out the mass, M = ρV = ρπh(R22 – R12), gives the desired result.
b. For a solid cylinder, R1 = 0, so I = ½ MR02.

44. Solution:

45. Determining Moments of Inertia
The parallel-axis theorem gives the moment of inertia about any axis parallel to an axis that goes through the center of mass of an object:

46. Determining Moments of Inertia
Parallel axis.
Determine the moment of inertia of a solid cylinder of radius R0 and mass M about an axis tangent to its edge and parallel to its symmetry axis.
Figure
Solution: ICM = ½ MR02, so I = 3/2 MR02.

47. Solution:

48. Determining Moments of Inertia
The perpendicular-axis theorem is valid only for flat objects.
Figure Caption: The perpendicular-axis theorem.

49. Rotational Kinetic Energy
The kinetic energy of a rotating object is given by
By substituting the rotational quantities, we find that the rotational kinetic energy can be written:
A object that both translational and rotational motion also has both translational and rotational kinetic energy:

50. Rotational Kinetic Energy
When using conservation of energy, both rotational and translational kinetic energy must be taken into account.
All these objects have the same potential energy at the top, but the time it takes them to get down the incline depends on how much rotational
inertia they have.
Figure

51. Rotational Kinetic Energy
The torque does work as it moves the wheel through an angle θ:

52. Rotational Plus Translational Motion; Rolling
In (a), a wheel is rolling without slipping. The point P, touching the ground, is instantaneously at rest, and the center moves with velocity .
In (b) the same wheel is seen from a reference frame where C is at rest. Now point P is moving with velocity – .
The linear speed of the wheel is related to its angular speed:
Figure Caption: (a) A wheel rolling to the right. Its center C moves with velocity v. Point P is at rest at this instant. (b) The same wheel as seen from a reference frame in which the axle of the wheel C is at rest—that is, we are moving to the right with velocity v. relative to the ground. Point P, which was at rest in (a), here in (b) is moving to the left with velocity –v as shown. (See also Section 3-9 on relative velocity.)

53. Why Does a Rolling Sphere Slow Down?
A rolling sphere will slow down and stop rather than roll forever. What force would cause this?
If we say “friction”, there are problems:
The frictional force has to act at the point of contact; this means the angular speed of the sphere would increase.
Gravity and the normal force both act through the center of mass, and cannot create a torque.
Figure Sphere rolling to the right.

54. Why Does a Rolling Sphere Slow Down?
The solution: No real sphere is perfectly rigid. The bottom will deform, and the normal force will create a torque that slows the sphere.
Figure The normal force, FN, exerts a torque that slows down the sphere. The deformation of the sphere and the surface it moves on has been exaggerated for detail.

55. Objects Rotating About a Fixed Axis
The rotational analog of linear momentum is angular momentum, L:
Then the rotational analog of Newton’s second law is:
This form of Newton’s second law is valid even if I is not constant.

56. Objects Rotating About a Fixed Axis
In the absence of an external torque, angular momentum is conserved:
More formally, the total angular momentum of a rotating object remains constant if the net external torque acting on it is zero.

57. Objects Rotating About a Fixed Axis
This means:
Therefore, if an object’s moment of inertia changes, its angular speed changes as well.
Figure 11-1: A skater doing a spin on ice, illustrating conservation of angular momentum: (a) I is large and ω is small; (b) I is smaller so ω is larger.
Figure 11-2: A diver rotates faster when arms and legs are tucked in than when they are outstretched. Angular momentum is conserved.

58. Angular Momentum of a Particle
A particle’s angular momentum.
What is the angular momentum of a particle of mass m moving with speed v in a circle of radius r in a counterclockwise direction?
Figure Caption: The angular momentum of a particle of mass m rotating in a circle of radius r with velocity v is L = r x mv.
Solution: Since L = r x p, L = mvr = mr2ω = Iω.

59. Angular Momentum of a Particle
The angular momentum of a particle about a specified axis is given by:
Figure The angular momentum of a particle of mass m is given by L = r x p = r x mv.

60. Angular Momentum of a Particle
If we take the derivative of , we find:
Since
we have:

61. Objects Rotating About a Fixed Axis
Object rotating on a string of changing length.
A small mass m attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a speed v1 = 2.4 m/s in a circle of radius R1 = 0.80 m. The string is then pulled slowly through the hole so that the radius is reduced to R2 = 0.48 m. What is the speed, v2, of the mass now?
Solution: Conservation of angular momentum gives the speed as 4.0 m/s.

62. Solution:

63. Objects Rotating About a Fixed Axis
Clutch.
A simple clutch consists of two cylindrical plates that can be pressed together to connect two sections of an axle, as needed, in a piece of machinery. The two plates have masses MA = 6.0 kg and MB = 9.0 kg, with equal radii R0 = 0.60 m. They are initially separated. Plate MA is accelerated from rest to an angular velocity ω1 = 7.2 rad/s in time Δt = 2.0 s. Calculate (a) the angular momentum of MA, and (b) the torque required to have accelerated MA from rest to ω1. (c) Next, plate MB, initially at rest but free to rotate without friction, is placed in firm contact with freely rotating plate MA, and the two plates both rotate at a constant angular velocity ω2, which is considerably less than ω1. Why does this happen, and what is ω2?
Solution: a. The angular momentum is 7.8 kg·m2/s.
b. The torque is the change in angular momentum divided by the time, 3.9 m·N.
c. Angular momentum is conserved (this is a rotational collision), so the new angular speed is 2.9 rad/s.

64. Objects Rotating About a Fixed Axis
Neutron star.
Astronomers detect stars that are rotating extremely rapidly, known as neutron stars. A neutron star is believed to form from the inner core of a larger star that collapsed, under its own gravitation, to a star of very small radius and very high density. Before collapse, suppose the core of such a star is the size of our Sun (r ≈ 7 x 105 km) with mass 2.0 times as great as the Sun, and is rotating at a frequency of 1.0 revolution every 100 days. If it were to undergo gravitational collapse to a neutron star of radius 10 km, what would its rotation frequency be? Assume the star is a uniform sphere at all times, and loses no mass.
Solution: Angular momentum is conserved; the rotation rate would be about 600 rev/s.

65. Solution:

66. Objects Rotating About a Fixed Axis
Angular momentum is a vector; for a symmetrical object rotating about a symmetry axis it is in the same direction as the angular velocity vector.
Figure (a) A person on a circular platform, both initially at rest, begins walking along the edge at speed v. The platform, assumed to be mounted on friction-free bearings, begins rotating in the opposite direction, so that the total angular momentum remains zero, as shown in (b).

67. Objects Rotating About a Fixed Axis
Running on a circular platform.
Suppose a 60-kg person stands at the edge of a 6.0-m-diameter circular platform, which is mounted on frictionless bearings and has a moment of inertia of 1800 kg·m2. The platform is at rest initially, but when the person begins running at a speed of 4.2 m/s (with respect to the Earth) around its edge, the platform begins to rotate in the opposite direction. Calculate the angular velocity of the platform.
Solution: Angular momentum is conserved, so the angular velocity is 0.42 rad/s.

68. Objects Rotating About a Fixed Axis
Spinning bicycle wheel.
Your physics teacher is holding a spinning bicycle wheel while he stands on a stationary frictionless turntable. What will happen if the teacher suddenly flips the bicycle wheel over so that it is spinning in the opposite direction?
Angular momentum is conserved, so the teacher will start spinning in the direction the wheel was spinning originally.

69. Angular Momentum and Torque for a System of Particles
The angular momentum of a system of particles can change only if there is an external torque—torques due to internal forces cancel.
This equation is valid in any inertial reference frame. It is also valid for the center of mass, even if it is accelerating:

70. Angular Momentum and Torque for a Rigid Object
For a rigid object, we can show that its angular momentum when rotating around a particular axis is given by:
Figure Calculating Lω = Lz = ΣLiz. Note that Li is perpendicular to ri and Ri is perpendicular to the z axis, so the three angles marked φ are equal.

71. Angular Momentum and Torque for a Rigid Object
Atwood’s machine.
An Atwood machine consists of two masses, mA and mB, which are connected by an inelastic cord of negligible mass that passes over a pulley. If the pulley has radius R0 and moment of inertia I about its axle, determine the acceleration of the masses mA and mB, and compare to the situation where the moment of inertia of the pulley is ignored.
Solution: First find the angular momentum of the system, and then apply the torque law. The torque is dL/dt, and a is dv/dt; taking the derivative of L and solving for a gives the solution in the text.

72. Angular Momentum and Torque for a Rigid Object
Bicycle wheel.
Suppose you are holding a bicycle wheel by a handle connected to its axle. The wheel is spinning rapidly so its angular momentum points horizontally as shown. Now you suddenly try to tilt the axle upward (so the CM moves vertically). You expect the wheel to go up (and it would if it weren’t rotating), but it unexpectedly swerves to the right! Explain.
Figure When you try to tilt a rotating bicycle wheel vertically upward, it swerves to the side instead.
Solution: You are trying to change the angular momentum vector (in direction, not magnitude). This requires a torque that points to the right (which is created by the upward force you exert on the wheel).

73. Angular Momentum and Torque for a Rigid Object
A system that is rotationally imbalanced will not have its angular momentum and angular velocity vectors in the same direction. A torque is required to keep an unbalanced system rotating.
Figure In this system L and ω are not parallel. This is an example of rotational imbalance.

74. Angular Momentum and Torque for a Rigid Object
Torque on unbalanced system.
Determine the magnitude of the net torque τnet needed to keep the illustrated system turning.
Solution: The net torque is the change in the angular momentum and equals ωL cos φ. The angular momentum is Iω/sin φ, so the torque is Iω2/tan φ.

75. Solution: θ ϕ Iω L

76. Conservation of Angular Momentum
If the net torque on a system is constant,
The total angular momentum of a system remains constant if the net external torque acting on the system is zero.

77. Conservation of Angular Momentum
Bullet strikes cylinder edge.
A bullet of mass m moving with velocity v strikes and becomes embedded at the edge of a cylinder of mass M and radius R0. The cylinder, initially at rest, begins to rotate about its symmetry axis, which remains fixed in position. Assuming no frictional torque, what is the angular velocity of the cylinder after this collision? Is kinetic energy conserved?
Figure Bullet strikes and becomes embedded in cylinder at its edge.
Solution: Angular momentum is conserved; kinetic energy is not. See text for detailed solution.

78. The Spinning Top and Gyroscope
A spinning top will precess around its point of contact with a surface, due to the torque created by gravity when its axis of rotation is not vertical.

79. The Spinning Top and Gyroscope

80. The Spinning Top and Gyroscope
The angular velocity of the precession is given by:
This is also the angular velocity of precession of a toy gyroscope, as shown.

81. Rotating Frames of Reference; Inertial Forces
There is an apparent outward force on objects in rotating reference frames; this is a fictitious force, or a pseudoforce. The centrifugal “force” is of this type; there is no outward force when viewed from an inertial reference frame.

82. The Coriolis Effect
The Coriolis effect is responsible for the rotation of air around low-pressure areas— counterclockwise in the Northern Hemisphere and clockwise in the Southern. The Coriolis acceleration is:
Figure (a) Winds (moving air masses) would flow directly toward a low-pressure area if the Earth did not rotate. (b) and (c): Because of the Earth’s rotation, the winds are deflected to the right in the Northern Hemisphere (as in Fig ) as if a fictitious (Coriolis) force were acting.

83. Summary Angles are measured in radians; a whole circle is 2π radians.
Angular velocity is the rate of change of angular position.
Angular acceleration is the rate of change of angular velocity.
The angular velocity and acceleration can be related to the linear velocity and acceleration.
The frequency is the number of full revolutions per second; the period is the inverse of the frequency.

84. Summary
The equations for rotational motion with constant angular acceleration have the same form as those for linear motion with constant acceleration.
Torque is the product of force and lever arm.
The rotational inertia depends not only on the mass of an object but also on the way its mass is distributed around the axis of rotation.
The angular acceleration is proportional to the torque and inversely proportional to the rotational inertia.

85. Summary
An object that is rotating has rotational kinetic energy. If it is translating as well, the translational kinetic energy must be added to the rotational to find the total kinetic energy.

86. Summary Angular momentum of a rigid object: Newton’s second law:
Angular momentum is conserved.
Torque:

87. Summary Angular momentum of a particle: Net torque:
If the net torque is zero, the vector angular momentum is conserved.